Given that 60.0 mL of 0.550 M NH4+ is titrated with 0.550 M NaOH. The Ka for NH4+ is 5.60x10^-10. Calculate the pH of the solution after 40.0 mL of NaOH has been added to the solution.

Answer 1

millimoles NH4^+ initially = mL x M = 60 x 0.550 = 33
millimoles NaOH added = 40 x 0.550 = 22
...............NH4^+ + NaOH ==> NH3 + H2O + Na^+
Initial........33..............0..............0.........0...........0
add.............................22.......................................
change...-22..............-22...........+22...................
equil..........11...............0..............22............................
Insert the equilibrium line into the Henderson-Hasslebalch equation.
pH = pKa + log [(base)/(acid)]
pH = + log (22/11) = ? or if you aren't familiar with the H-H equation, from the ionization of NH4^+ you get
...................NH4^+ ==> NH3 + H^+
Ka = (NH3)(H^+)/(NH4^+). Solve for
(H^+) = Ka(NH4+)/(NH3)
(H^+) = 5.6E-10*(11/22)
Then convert to pH.
Post your work if you get stuck.

Answer 2

Okay, thanks. That helped me a lot. How would you go about the same problem after 60.0mL of NaOH has been added to the solution? I think I know how to do it but I am not sure how to set up the equilibrium equation